Chapter 11 Additional Topics

SSM: Intermediate Algebra Homework 11.8
x
2
y = − x + 6
2 2
− 2 = − x + 6
2
2x
= 8
x
2
= 4
x = ± 2
Let x = − 2 and x = 2 in
for y.
( ) 2
y = −2 − 2
= 4 − 2
= 2
( ) 2
y = 2 − 2
= 4 − 2
= 2
2, 2
2
y = x − 2 and solve
The solutions are ( − ) and ( )
7. x
2 y
2
x
+ = 49
2 2
+ y = 16
y
2, 2 .
x
2
2 2
2
2 2
( x )
2
x
+ y = 25
+ − − 1 = 25
x + x + 2x
+ 1 = 25
2x
+ 2x
− 24 = 0
x
2
+ x − 12 = 0
( x )( x )
+ 4 − 3 = 0
x + 4 = 0 or x − 3 = 0
x = − 4 or x = 3
Let x = − 4 and x = 3 in y = −x
− 1 and solve
for y.
y = − −4 −1
y = − 3 −1
( )
= 4 −1
= 3
( )
= −3 −1
= −4
4,3
The solutions are ( − ) and ( 3, 4)
− .
−8
8
−8
8
x
11. y
2 x
2 2
− = 16
y + x = 4
4
y
(0, 4)
The graphs do not intersect. The solution set is
the empty set, ∅ .
Solve using elimination.
x
2 2
+ y = 49
2 2
−x
− y = −16
0 = 33 False
There is no solution.
9. x
2 + y
2 = 25
(−4, 3)
−4
y = −x
−1
4
−4
y
4
x
(3, −4)
The two intersection points ( − 4,3)
and ( 3, − 4)
are the solutions of the system.
Solve using substitution.
Substitute −x
− 1 for y in the first equation.
−4
(−3, −5)
−2
4
(3, −5)
The three intersection points ( −3, − 5)
, ( )
and ( 3, − 5)
are the solutions to the system.
Solve using elimination.
y
2 2
2
− x = 16
2
y + x = 4
y
+ y = 20
y
2
+ y − 20 = 0
( y )( y )
+ 5 − 4 = 0
y + 5 = 0 or y − 4 = 0
y = − 5 or y = 4
x
2
0, 4 ,
Let y = − 5 and y = 4 in y + x = 4 and solve
for x.
2
2
− 5 + x = 4 4 + x = 4
x
2
= 9
x = ± 3
x
2
= 0
x = 0
3, 5
The solutions are ( − − ) , ( 3, − 5)
, and ( )
0, 4 .
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